poj 2379 Sum of Consecutive Prime Numbers
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 24498 | Accepted: 13326 |
Description
numbers, so neither 7 + 13 nor 3 + 5 + 5 + 7 is a valid
representation for the integer 20.
Your mission is to write a program that
reports the number of representations for the given positive integer.
Input
separate line. The integers are between 2 and 10 000, inclusive. The end of the
input is indicated by a zero.
Output
corresponding to an input line except the last zero. An output line includes the
number of representations for the input integer as the sum of one or more
consecutive prime numbers. No other characters should be inserted in the
output.
Sample Input
2
3
17
41
20
666
12
53
0
Sample Output
1
1
2
3
0
0
1
2 题意:某些数可以是连续质数的和,那么给定数a,求a的分解有几种可能。
思路:先可以用埃氏筛选找出连续的质数,之后尺取法找可能的情况,值得注意的是两组可能的连续质数序列存在元素重叠的情况,所以每找到一组,尺取法的头部和尾部统一更新为上一种情况的尾部再加1,继续查找下一种情况。。。
AC代码:
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
using namespace std;
const int N_MAX = + ;
int prime[N_MAX],res;
bool is_prime[N_MAX+],what;
int sieve(int n) {
int p = ;
for (int i = ; i <= n; i++)is_prime[i] = true;
is_prime[] = is_prime[] = false;
for (int i = ; i <= n; i++) {
if (is_prime[i]) {
prime[p++] = i;
for (int j = i*i; j <= n; j += i)is_prime[j] = false;
}
}
return p;
}
int main() {
int n=sieve(N_MAX),a;
while (scanf("%d",&a)&&a) {
res = ,what = ;
int l= , r= , sum = ;
int bound=upper_bound(prime, prime + n, a)-prime;
while () {////////
for (;;) {////
while (r < bound&&sum < a) {
sum += prime[r++];
}
if (sum == a) {
res++;
what = ;
break;
}
if (sum < a) {//找不到了,终止搜索;
what = ;
break;
}
sum -= prime[l++];
}////
if (what) { l++; r = l; sum = ; }
else break;
}////////
printf("%d\n", res);
}
return ;
}
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